Assume there is a known probability that it's going to rain today, let's call it P
.
If it rains today, the probability of rain tomorrow is P + step
.
If it doesn't rain today, the probability of rain tomorrow is P - step
.
Once the probability of rain reaches 1.0
, the eternal rain starts. If it reaches 0.0
, the dry spell falls on Probland. There are N+1
possible states, step = 1/N
Question:
Given that you start in state Si
(probability of rain today is Pi
) and the step by which you increase/decrease the probability of rain tomorrow is step = 1/N
, what's the probability of eventually reaching state Sn
with Pn = 1.0
?
Illustration:
P0 = 0/4 = 0
P1 = 1/4
P2 = 2/4 = 1/2
P3 = 3/4
P4 = 4/4 = 1
The probability of eternal rain in S4
is P4 = 4/4 = 1
. In S4
, it rains today with probability 1 and it's going to rain tomorrow with probability 1 and so forth. By the same logic, P0 = 0
Analytical solution:
The probabilty of eternal rain if we start in Si
is Pri
Pr4 = 1
Pr3 = P3 * Pr4 + (1 - P3) * Pr2
Pr2 = P2 * Pr3 + (1 - P2) * Pr1
Pr1 = P1 * Pr2 + (1 - P1) * Pr0
Pr0 = 0
Solving this system of equations, we get Pr0 = 0, Pr1 = 1/8, Pr2 = 1/2, Pr3 = 7/8, Pr4 = 1
Simulated paths solution:
Start in a given state, simulate runs (all of them will terminate in 0 or 1), divide the sum of the results of all runs by the total number of runs. That ratio will converge to the theoretically computed one as the number of runs grows.